Evaluate the definite integral: ∫ 0 1 x x 2 + 1 d x

Let I = ∫ 0 1 x d x x 2 + 1 Put x 2 = t ⇒ 2 x d x = d t When x = 0 ⇒ t = 0 and When x = 1 ⇒ t = 1 ∴ I = ∫ 0 1 d t 2 ( t + 1 ) I = 1 2 log | t + 1 | 0 1 I = 1 2 log 2 - log 1 I = 1 2 log 2

Evaluate the definite integral: ∫ 0 1 x x 2 + 1 d x

Let I = ∫ 0 1 x d x x 2 + 1 Put x 2 = t ⇒ 2 x d x = d t When x = 0 ⇒ t = 0 and When x = 1 ⇒ t = 1 ∴ I = ∫ 0 1 d t 2 ( t + 1 ) I = 1 2 log | t + 1 | 0 1 I = 1 2 log 2 - log 1 I = 1 2 log 2

Evaluate the following definite integrals: ∫ 0 π 3 cos x 3 + 4 sin x d x .

Let I = ∫ 0 π 3 cos x 3 + 4 sin x d x Put sin x = t ⇒ cos x d x = d t When x = π 3 ⇒ t = 3 2 I = ∫ 0 3 2 d t 3 + 4 t I = log | 3 + 4 t | 4 0 3 2 I = 1 4 log | 3 + 2 3 | - log 3 I = 1 4 log 3 + 2 3 3

Evaluate the definite integral: ∫ 1 2 3 x 9 x 2 - 1 d x

Let I = ∫ 1 2 3 x 9 x 2 - 1 d x I = 1 6 ∫ 1 2 18 x 9 x 2 - 1 ∵ ∫ f ' x f x d x = log | f x | I = 1 6 log | 9 x 2 - 1 | 1 2 I = 1 6 log | 36 - 1 | - log | 9 - 1 | I = 1 6 log 35 - log 8 I = 1 6 log 35 8

Evaluate the following definite integrals: ∫ a b log x x d x

Put log x = t ⇒ 1 x d x = d t When x = a ⇒ t = log a When x = b ⇒ t = log b ∴ I = ∫ log a log b t d t I = t 2 2 log a log b I = 1 2 log b 2 - log a 2 I = 1 2 log b - log a log b + log a I = 1 2 log b a log a b

Evaluate the definite integral: ∫ 0 2 6 x + 3 x 2 + 4 d x

Let I = ∫ 0 2 6 x + 3 x 2 + 4 d x I = ∫ 0 2 6 x x 2 + 4 d x + 3 2 tan - 1 x 2 0 2 Put x 2 = t in first integral ∴ 2 x d x = d t When x = 0 ⇒ t = 0 and When x = 2 ⇒ t = 4 ∴ I = 3 ∫ 0 4 d t t + 4 + 3 2 × π 4 I = 3 log | t + 4 | 0 4 + 3 π 8 I = 3 log 8 - log 4 + 3 π 8 I = 3 log 2 + 3 π 8

Evaluate ∫ 0 π / 2 1 1 + sin x d x .

Let I = ∫ 0 π 2 d x 1 + sin x Put tan x 2 = t ⇒ s e c 2 x 2 1 2 d x = d t d x = 2 d t 1 + t 2 and sin x = 2 t 1 + t 2 When x = 0 ⇒ t = 0 and When x = π ⇒ t → ∞ I = ∫ 0 ∞ 2 d t 1 + t 2 1 + 2 t 1 + t 2 I = L t b → ∞ ∫ 0 b 2 d t 1 + t 2 + 2 t I = L t b → ∞ ∫ 0 b 2 d t 1 + t 2 I = L t b → ∞ - 1 1 + t 0 b I = - 2 [ 0 - 1 ] I = 2

Evaluate ∫ 0 1 x 5 1 + x 6 d x

Let I = ∫ 0 1 x 5 1 + x 6 d x Put x 6 = t ⇒ 6 x 5 d x = d t When x = 0 ⇒ t = 0 When x = 1 ⇒ t = 1 ∴ I = ∫ 0 1 d t 6 1 + t I = 1 6 log 1 + t 0 1 I = 1 6 ( log 2 - log 1 ) = 1 6 log 2

Evaluate ∫ 0 1 x 1 + x 2 d x .

Let I = ∫ 0 1 x 1 + x 2 d x I = 1 2 ∫ 0 1 ( 1 + x 2 ) - 1 2 · 2 x · d x I = 1 2 ( 1 + x 2 ) 1 2 1 2 0 1 I = 2 - 1 I = 2 - 1

Evaluate the following definite integrals: ∫ 0 4 d x x 2 + 2 x + 3

Let I = ∫ 0 4 d x x 2 + 2 x + 3 I = ∫ 0 4 d x x 2 + 2 x + 1 + 2 I = ∫ 0 4 d x x + 1 2 + 2 2 Put x + 1 = t ⇒ d x = d t When x = 0 ⇒ t = 1 When x = 4 ⇒ t = 5 I = ∫ 0 5 d x t 2 + 2 2 I = log | t + t 2 + 2 | 1 5 I = log | 5 + 25 + 2 | - log | 1 + 1 2 + 2 | I = log 5 + 27 1 + 3 I = log 5 + 2 3 1 + 3